Friction Problem -- String is tied to a block and a hanging bucket

[Nithya115]

USE OF THE PHYSICS FORUMS TEMPLATE IS REQUIRED ON ALL HOMEWORK PROBLEMS. PLEASE COMPLY WITH THIS IN THE FUTURE.

A string is tied to a 4.4 kg block and 120g hanging bucket. Students add 20g washers one at a time to the bucket. The student are unaware that the
coefficient of static friction for the block on the table is 0.42.
A) what is the maximum force of static friction for the block?
Answer: 18.11N
B) how many washers can the students add to the bucket without moving the block?
Answer: 86
C) the coefficient on kinetic friction is 0.34. Calculate the acceleration of the block when the final washer is added to the bucket and the objects start to move...

How do get c)? I got 0.56m/s^2 but the answer in the book is 0.75m/s^2

 

[Chestermiller]

Is it correct to assume that there is a pulley present somewhere?

Chet

 

[Nithya115]

Is it correct to assume that there is a pulley present somewhere?

Chet

ya there is a pulley between the bucket and the block. Basically, the block is sitting on the table and there is a pulley at the edge of the table. The bucket is hanging from the string that is on the pulley. its like the usual diagrams.

I also did the same procedure but I am not getting that answer!

 

[Chestermiller]

Show us how you got your answer? How can we help you if we.don't know the details of what you did?

Chet

 

[Nithya115]

ya there is a pulley between the bucket and the block. Basically, the block is sitting on the table and there is a pulley at the edge of the table. The bucket is hanging from the string that is on the pulley. its like the usual diagrams.

Show us how you got your answer? How can we help you if we.don't know the details of what you did?

Chet

a= m2g -Fs/ m1 + m2
a= (1.86 x 9.8) - 15/ (4.4 + 1.86)
a= 0.52m/s^2I got Fs by...
Fs = (0.34)(Fn=4.4x9.8)
= 1.86

I got m2 by...
1. calculating the mass of the washers
20g x 87 washers = 1740g
2. adding the bucket and washers
120g + 1740g = 1860g
3. converted to kg
1860g/1000g=1.86 kg

 

[Chestermiller]

Your methodology is solid. I get 0.57 m/s², in rough agreement with your 0.56. The book shows 0.75, but they got the digits interchanged. For the kinetic friction force, I got 14.7 N.

Chet

 

[Brul]

Just a second confirmation; I get 0.58 m/s² as well.

 

[Chestermiller]

Just a second confirmation; I get 0.58 m/s² as well.

Hi Brul. Welcome to Physics Forums.

Chet

 

[Brul]

Hello and thank you!

 

[Nithya115]

Your methodology is solid. I get 0.57 m/s², in rough agreement with your 0.56. The book shows 0.75, but they got the digits interchanged. For the kinetic friction force, I got 14.7 N.

Chet

Thanks. I rounded the kinetic force to 15 which I probably should avoid.